A mistake here (a≥2)

welcome to AOPS @daixiahu!!!! already heping out on your first day!!

if $s\nmid r$,let$r=sp+q$,$0<q<s$, $p,q\in \mathbb{Z}$,then $a^r+b^r=a^{sp+q}+b^{sp+q}=a^{sp}\cdot a^q+b^{sp}\cdot b^q$ $=(a^q+b^q)(a^{sp}+b^{sp})-(a^{sp}\cdot b^q+b^{sp}\cdot a^q)$ since $a^s+b^s\mid a^r+b^r$and $a^s+b^s\mid (a^q+b^q)(a^{sp}+b^{sp})$, we have $a^s+b^s\mid (a^{sp}\cdot b^q+b^{sp}\cdot a^q)\mid a^s+b^s$ $\Rightarrow (a^{sp}\cdot a^q+b^{sp}\cdot b^q)+(a^{sp}\cdot b^q+b^{sp}\cdot a^q)=(a^q+b^q)(a^{sp}+b^{sp})$ $\Rightarrow a^s+b^s\mid a^q+b^q\Rightarrow a^s+b^s\le a^q+b^q\Rightarrow s\le q$,contradiction with $q<s$. $\square$